A5-S-431F08 - SOLUTIONS TO ASSIGNMENT 5 ACTSC 431/831, FALL...

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Unformatted text preview: SOLUTIONS TO ASSIGNMENT 5 ACTSC 431/831, FALL 2008 1. (a) We have E ( S ) = E ( N ) E ( X ) = 10 and Pr { S = 1 } = Pr { S = 3 } = 0 , Pr { S = 0 } = Pr { N = 0 } = 0 . 1 , Pr { S = 2 } = Pr { N = 1 ,X 1 = 2 } = 0 . 04 , Pr { S = 4 } = Pr { N = 1 ,X 1 = 4 } + Pr { N = 2 ,X 1 = 2 ,X 2 = 2 } = 0 . 128 . Thus, we have E [( S- 5) + ] = E ( S )- E ( S 5) = 10- 4 X x =1 x Pr { S = x } - 5 1- 4 X x =0 Pr { S = x } ! = 5 . 748 . (b) We have E [( S- 4 . 6) + ] = E ( S )- E ( S 4 . 6) = 10- 4 X x =1 x Pr { S = x } - 4 . 6 1- 4 X x =0 Pr { S = x } ! = 6 . 041 . (c) We have E [( S- 4) + ] = E ( S )- E ( S 4) = 10- 3 X x =1 x Pr { S = x } - 4 1- 3 X x =0 Pr { S = x } ! = 6 . 48 , V ar ( S ) = E ( N ) V ar ( X ) + V ar ( N )( E ( X )) 2 = 40 , and E [( S- 4) 2 + ] = X x =5 ( x- 4) 2 Pr { S = x } = E [( S- 4) 2 ]- 4 X x =0 ( x- 4) 2 Pr { S = x } = V ar ( S ) + ( E ( S )) 2- 8 E ( S ) + 16- 4 X x =0 ( x- 4) 2 Pr { S = x } = 74 . 24 . Hence, V ar [( S- 4) + ] = E [( S- 4) 2 + ]- ( E [( S- 4) + ]) 2 = 32 . 25. 2. (a) We have E ( N ) = 15 and E [( X i- 25) + ] = E ( X i )- E ( X i 25) = 20- Z 25 e- x/ 20 dx = 20 e- 5 / 4 = 5 . 7301 . Thus, the net reinsurance premium for the excess-of-loss reinsurance is E ( N ) E [( X i- 25) + ] = 85 . 95 . 1 (b) The survival function of S is 1- F S ( x ) = 15 16 e- x 320 , x . Thus, the net reinsur- ance premium for the stop-loss reinsurance is E [( S- 150) + ] = E ( S )- E ( S 150) = E ( N ) E ( X i )- Z 150 15 16 e- x/ 320 dx = 300 e- 15 / 32 = 187 . 735 ....
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A5-S-431F08 - SOLUTIONS TO ASSIGNMENT 5 ACTSC 431/831, FALL...

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